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3 years ago

I need help with this work

Physics

2 answers:

san4es73 [151]3 years ago

7 0

What work??? I don’t see anything

8_murik_8 [283]3 years ago

3 0

Look up a net force calculator :)

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5. Calculate the final speed of a 110-kg rugby player who is initially running at 8.00 m/s but collides head-on with a padded go

bulgar [2K]

Answer:

0.8 m/s

Explanation:

From Newton's third law of motion,

Impulse on a body = Change in momentum of the body

Ft = m(v-u) ......................... Equation 1

Where F = force on the body , t = time, m = mass of the body, v = final velocity, u = initial velocity,

Note: Taking the direction of the initial velocity as negative,

Given: F = 1.76×10⁴ N, t = 5.5×10⁻² s, m = 110 kg, u = -8.0 m/s

Substitute into equation 1

1.76×10⁴ ( 5.5×10⁻²) = 110[v-(-8.0)]

968 = 110v+880

110v = 968-880

110v = 88

v = 88/110

V = 0.8 m/s

Hence the final velocity = 0.8 m/s

4 0

4 years ago

A person hears a siren as a fire truck approaches and passes by. The frequency varies from 480Hz on approach to 400Hz going away

alekssr [168]

Answer:

31.2 m/s

Explanation:

$f_{app}$ = Frequency of approach = 480 Hz

$f_{aw}$ = Frequency of going away = 400 Hz

$V$ = Speed of sound in air = 343 m/s

$v$ = Speed of truck

Frequency of approach is given as

$f_{app} = \frac{Vf}{V - v}$ eq-1

Frequency of moving awayy is given as

$f_{aw} = \frac{Vf}{V + v}$ eq-2

Dividing eq-1 by eq-2

$\frac{f_{app}}{f_{aw}} = \frac{V + v}{V - v}$

$\frac{480}{400} = \frac{343 + v}{343 - v}$

$v$ = 31.2 m/s

7 0

3 years ago

Find a numerical value for rhoearth, the average density of the earth in kilograms per cubic meter. Use 6378km for the radius of

mylen [45]

Answer:

density = 5520 kg/m^3

Explanation:

given that

radius of earth = 6378 km

G = 6.67 x 10⁻¹¹ m³/kg.s²

g = 9.80 m/s²

we know,

$g = \dfrac{GM}{r^2}$

mass of earth

$M = \dfrac{gr^2}{G}$

$M = \dfrac{9.8 \times (6378 \times 10^3)^2}{6.67 \times 10^{-11}}$

M = 5.972 x 10²⁴ kg

density = $\dfrac{mass}{volume}$

V = volume of the earth = 4/3πr³

V = 4/3 x 3.14 x (6378 x 10³)³

V = 1.08 x 10²¹ m³

density = $\dfrac{5.972\times 10^{24}}{1.08\times 10^{21}}$

density = 5.52 x 10³ kg/m^3

density = 5520 kg/m^3

8 0

3 years ago

A 3 kg mass object is pushed 0.6 m into a spring with spring constant 210 N/m on a frictionless horizontal surface. Upon release

Svetradugi [14.3K]

Answer with Explanation:

We are given that

Mass of spring,m=3 kg

Distance moved by object,d=0.6 m

Spring constant,k=210N/m

Height,h=1.5 m

a.Work done to compress the spring initially= $\frac{1}{2}kx^2=\frac{1}{2}(210)(0.6)^2=37.8J$

By conservation law of energy

Initial energy of spring=Kinetic energy of object

$37.8=\frac{1}{2}(3)v^2$

$v^2=\frac{37.8\times 2}{3}$

$v=\sqrt{\frac{37.8\times 2}{3}}$

v=5.02 m/s

c.Work done by friction on the incline, $w_{friction}=P.E-spring \;energy$

$W_{friction}=3\times 9.8\times 1.5-37.8=6.3 J$

8 0

3 years ago

2.) How much momentum does a stationary 5500 kg mass have?

EleoNora [17]

momentum is measured by the product of mass and velocity of the object. As You have mentioned that the body is stationary then the velocity is zero. So the momentum of the body is 0.

hope this helps

3 0

3 years ago

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